Representing and Analyzing SHM - AP Physics C Mechanics

  1. Find the time period for a pendulum with L=1L = 1 m and g=9.8g = 9.8 m/s².
    T=2π19.8 sT = 2\pi \sqrt{\frac{1}{9.8}} \text{ s}. T=2π19.8=2π0.102≈2.0T = 2\pi\sqrt{\frac{1}{9.8}} = 2\pi\sqrt{0.102} \approx 2.0 s.
  2. What is the restoring force in a spring-mass system?
    F=−kxF = -kx. Hooke's law: force is proportional to displacement and directed toward equilibrium.
  3. What is the expression for velocity as a function of time in SHM?
    v(t)=−Aωsin⁡(ωt+ϕ)v(t) = -A\omega \sin(\omega t + \phi). Velocity is the time derivative of displacement x(t)x(t).
  4. Calculate the potential energy at maximum displacement.
    Equal to total mechanical energy: E=12kA2E = \frac{1}{2}kA^2. At maximum displacement, all energy is potential energy.
  5. Describe how amplitude affects the period in SHM.
    Amplitude does not affect the period. Period depends only on system parameters mm and kk, not amplitude.
  6. What is the formula for the displacement at time tt in SHM?
    x(t)=Acos⁡(ωt+ϕ)x(t) = A \cos(\omega t + \phi). Standard position function for simple harmonic motion.
  7. Identify the phase constant in the equation x(t)=A cos⁡(ωt+ϕ)x(t) = A \, \cos(\omega t + \phi).
    ϕ\phi. The phase constant determines initial conditions of the oscillation.
  8. What is the maximum velocity in SHM given amplitude AA and angular frequency ω\omega?
    vmax=Aωv_{max} = A\omega. Maximum speed occurs at equilibrium position where displacement is zero.
  9. State the formula for the angular frequency in SHM.
    ω=km\omega = \sqrt{\frac{k}{m}}. Derived from F=−kxF = -kx and Newton's second law for oscillatory motion.
  10. Calculate the frequency of oscillation given T=0.5T = 0.5 s.
    f=2 Hzf = 2 \text{ Hz}. Frequency is the reciprocal of period: f=1T=10.5=2f = \frac{1}{T} = \frac{1}{0.5} = 2 Hz.
  11. How does mass affect the period of a mass-spring system?
    Period increases with mass as T=2πmkT = 2\pi\sqrt{\frac{m}{k}}. Period increases with square root of mass.
  12. How is the natural frequency ω0\omega_0 defined for a simple pendulum?
    ω0=gL\omega_0 = \sqrt{\frac{g}{L}}. Natural frequency for small amplitude pendulum oscillations.
  13. What is the formula for the period of a mass-spring system?
    T=2πmkT = 2\pi \sqrt{\frac{m}{k}}. Derived by substituting ω=km\omega = \sqrt{\frac{k}{m}} into T=2πωT = \frac{2\pi}{\omega}.
  14. Identify the phase of motion when displacement is zero.
    Equilibrium position. Zero displacement corresponds to passing through equilibrium.
  15. What is the condition for resonance in mechanical systems?
    ω=ω0\omega = \omega_0. Resonance occurs when driving frequency equals natural frequency.
  16. How do you calculate the total mechanical energy in SHM?
    E=12kA2E = \frac{1}{2}kA^2. Total energy equals maximum potential energy at amplitude.
  17. Find the maximum acceleration in SHM with A=0.5A = 0.5 m and ω=2\omega = 2 rad/s.
    amax=2 m/s2a_{max} = 2 \text{ m/s}^2. Maximum acceleration occurs at amplitude: amax=Aω2=0.5×22=2a_{max} = A\omega^2 = 0.5 \times 2^2 = 2 m/s².
  18. Calculate the angular frequency for a mass-spring system with k=100k = 100 N/m and m=4m = 4 kg.
    ω=5 rad/s\omega = 5 \text{ rad/s}. ω=km=1004=25=5\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{100}{4}} = \sqrt{25} = 5 rad/s.
  19. What characterizes simple harmonic motion?
    Motion where restoring force is proportional to displacement. Defining characteristic: F=−kxF = -kx creates oscillatory motion.
  20. What is the kinetic energy at maximum displacement in SHM?
    Zero. At amplitude, velocity is zero so kinetic energy is zero.
  21. State the relationship between period TT and angular frequency ω\omega.
    T=2πωT = \frac{2\pi}{\omega}. Period is the time for one complete oscillation cycle.
  22. What is the expression for potential energy in a spring-mass system?
    U=12kx2U = \frac{1}{2}kx^2. Elastic potential energy stored in the compressed or stretched spring.
  23. Calculate the mechanical energy in SHM with k=50k = 50 N/m and A=0.2A = 0.2 m.
    E=1 JE = 1 \text{ J}. E=12kA2=12×50×(0.2)2=1E = \frac{1}{2}kA^2 = \frac{1}{2} \times 50 \times (0.2)^2 = 1 J.
  24. What is the relationship between frequency and period?
    f=1Tf = \frac{1}{T}. Frequency counts cycles per unit time, period is time per cycle.
  25. What is the expression for the velocity at equilibrium position?
    Maximum velocity: vmax=Aωv_{max} = A\omega. At equilibrium, kinetic energy is maximum and equals AωA\omega.
  26. Identify the formula for acceleration in SHM.
    a(t)=−ω2x(t)a(t) = -\omega^2 x(t). Acceleration is proportional to displacement but opposite in direction.
  27. Identify the unit of angular frequency.
    radians per second (rad/s). Angular frequency measures how fast the phase angle changes with time.
  28. Find the maximum acceleration in SHM with A=0.5 mA = 0.5 \, \text{m} and ω=2 rad/s\omega = 2 \, \text{rad/s}.
    amax=2 m/s2a_{max} = 2 \text{ m/s}^2. Maximum acceleration occurs at amplitude: amax=Aω2=0.5×22=2 m/s2a_{max} = A\omega^2 = 0.5 \times 2^2 = 2 \text{ m/s}^2.
  29. What is the kinetic energy formula in a spring-mass system?
    K=12mv2K = \frac{1}{2}mv^2. Standard kinetic energy formula for any moving object.
  30. Find the period of a pendulum with length LL and gravity gg.
    T=2πLgT = 2\pi \sqrt{\frac{L}{g}}. For small angles, pendulum exhibits simple harmonic motion.
  31. What determines the period of oscillation in a simple pendulum?
    Length of pendulum and gravitational acceleration. Period is independent of mass and amplitude for small oscillations.
  32. Calculate the frequency if angular frequency ω=10\omega = 10 rad/s.
    f=102π Hzf = \frac{10}{2\pi} \text{ Hz}. Frequency relates to angular frequency by f=ω2πf = \frac{\omega}{2\pi}.
  33. Find the expression for potential energy at any point in SHM.
    U=12kx2U = \frac{1}{2}kx^2. Potential energy varies with displacement squared.
  34. Describe the relationship between total energy and amplitude in SHM.
    E∝A2E \propto A^2. Total energy is proportional to amplitude squared.
  35. What is the general equation for displacement in SHM?
    x(t)=A cos⁡(ωt+ϕ)x(t) = A \, \cos(\omega t + \phi). General form where AA is amplitude, ω\omega is angular frequency, and ϕ\phi is phase constant.
  36. Find the displacement at t=0t = 0 if A=2A = 2 m and ϕ=π4\phi = \frac{\pi}{4}.
    x(0)=2 mx(0) = \sqrt{2} \text{ m}. At t=0t = 0: x(0)=2cos⁡(π4)=2×12=2x(0) = 2\cos(\frac{\pi}{4}) = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2} m.
  37. State the formula for the angular frequency of a simple harmonic oscillator.
    ω=km\omega = \sqrt{\frac{k}{m}}. Fundamental relationship for harmonic oscillator systems.
  38. Identify the term for the maximum displacement from equilibrium.
    Amplitude. Amplitude is the maximum distance from equilibrium position.
  39. What is the dimension of angular frequency?
    [T−1T^{-1}]. Angular frequency has dimension of inverse time.

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